HCF MCQ Quiz - Objective Question with Answer for HCF - Download Free PDF

Given:

p(x) = 2x³ – 3x² – 2x + 3 and q(x) = 3x² + 8x + 5

Concept:

H.C.F. of two or more equations is the greatest factor that divides each of them exactly.

Calculation:

The factors of p(x) = 2x³ – 3x² – 2x + 3

⇒ x² × (2x – 3) – 1 × (2x – 3)

⇒ (x² – 1) × (2x – 3)

⇒ (x – 1) × (x + 1) × (2x – 3)

And, the factors of q(x) = 3x² + 8x + 5

⇒ 3x² + 5x + 3x + 5

⇒ x × (3x + 5) + 1 × (3x + 5)

⇒ (3x + 5) × (x + 1)

Given:

Numbers: \(4\frac{1}{5}\), \(5\frac{2}{7}\), \(7\frac{4}{9}\)

Formula Used:

HCF of fractions = HCF of numerators / LCM of denominators

Calculation:

Convert mixed fractions to improper fractions:

\(4\frac{1}{5}\) = (4 × 5 + 1)/5 = 21/5

\(5\frac{2}{7}\) = (5 × 7 + 2)/7 = 37/7

\(7\frac{4}{9}\) = (7 × 9 + 4)/9 = 67/9

Extract numerators and denominators:

Numerators: 21, 37, 67

Denominators: 5, 7, 9

Find HCF of numerators:

HCF(21, 37, 67) = 1 (since all are co-prime)

Find LCM of denominators:

LCM(5, 7, 9):

Prime factorization:

5 = 5¹

7 = 7¹

9 = 3²

LCM = 5¹ × 7¹ × 3² = 315

HCF = HCF of numerators / LCM of denominators

HCF = 1 / 315

The HCF of \(4\frac{1}{5}\), \(5\frac{2}{7}\), and \(7\frac{4}{9}\) is 1/315 .

Given:

Oil quantities:

1st quality = 725 liters

2nd quality = 783 liters

3rd quality = 812 liters

Formula used:

Calculate the Highest Common Factor (HCF) of the given quantities to find the least possible number of bottles.

Calculation:

Find HCF of 725, 783, and 812.

Prime factorization:

725 = 5 × 5 × 29

783 = 3 × 3 × 3 × 29

812 = 2 × 2 × 7 × 29

Common factor = 29

Calculate the least possible number of bottles:

⇒ Total bottles = (725 ÷ 29) + (783 ÷ 29) + (812 ÷ 29)

⇒ Total bottles = 25 + 27 + 28 = 80

∴ The correct answer is option (4).

Given:

Number of Neem trees = 84

Number of Oak trees = 126

Number of Banyan trees = 168

Formula used:

To find the minimum number of rows required, calculate the HCF (Highest Common Factor) of the given numbers.

Minimum number of rows = HCF of 84, 126, and 168

Calculation:

Find the HCF of 84, 126, and 168

Prime factorization of 84 = 2 × 2 × 3 × 7

Prime factorization of 126 = 2 × 3 × 3 × 7

Prime factorization of 168 = 2 × 2 × 2 × 3 × 7

⇒ HCF = 2 × 3 × 7 = 42

Divide the total number of trees by the HCF to determine the number of rows for each type of tree.

Number of rows for Neem trees = 84 ÷ 42 = 2

Number of rows for Oak trees = 126 ÷ 42 = 3

Number of rows for Banyan trees = 168 ÷ 42 = 4

Add the rows for each type of tree.

Total rows = 2 + 3 + 4 = 9

∴ The correct answer is option (1).

Given:

Numbers are 68, 140, and 248.

We need to find the greatest number that divides these numbers leaving the same remainder.

Formula Used:

When a number divides multiple numbers leaving the same remainder, the required number is the HCF (Highest Common Factor) of the differences of these numbers.

Calculation:

Find the differences:

140 - 68 = 72

248 - 140 = 108

248 - 68 = 180

Now, find the HCF of 72, 108, and 180.

Prime factorization:

72 = 2³ × 3²

108 = 2² × 3³

180 = 2² × 3² × 5

Common factors: 2² × 3²

HCF = 4 × 9 = 36

The greatest number that divides 68, 140, and 248 leaving the same remainder is 36.

Given:

Length of timber₁ = 143 m

Length of timber2 = 78 m

Length of timber3 = 117 m

Calculation:

Greatest possible length of each plank = HCF of 143, 78 and 117

143 = 13 × 11

78 = 13 × 2 × 3

117 = 13 × 3 × 3 

HCF is 13

∴ Greatest possible length of each plank is 13 m.

Given:

The sum of two numbers is 288 and their HCF is 16

Calculations:

Let the ratio of number be x : y

So the numbers will be 16x & 16y (HCF is an integral part of a number)

According to the question

16x + 16y = 288

⇒ 16(x + y) = 288

⇒ x + y = 18

Pairs of x, y can be (1, 17) (5, 13) (7, 11)

So there can be only 3 pairs.

∴ The correct choice is option 1.

Given:

Ratio of numbers = 7 ∶ 11

HCF = 28

Calculation:

Let the numbers be 7x and 11x

HCF of 7x and 11x is x

HCF = x = 28

The numbers will be 7 × 28 and 11 × 28

⇒ The numbers will be 196 and 308

Sum of numbers = 196 + 308

⇒ Sum of numbers = 504

∴ Sum of numbers is 504

Shortcut Trick Note that the sum of two numbers is asked.

Let the numbers be 7x and 11x.

Add the numbers:

⇒ 7x + 11x

⇒ 18x

Now see, the final number must be the multiple of 18, so in options only 504 is multiple of 18.

∴ The sum of two number is 504.

Given:

(4315 − 1) and (425 − 1)

Concept used:

HCF of (a^m − 1) and (aⁿ − 1) is (a^HCF(m,n) − 1).

Calculations:

HCF(315, 25) = 5

According to the concept,

HCF {(4315 − 1), (425 − 1)} 

= (4^{HCF(315, 25)} − 1)

= (4⁵ − 1)

= 1024 − 1

= 1023

Hence, The Required value is 1023.

Given:

The H.C.F. of (x3 + x2 + x + 1) and (x4 – 1) is

Calculation:

⇒  (x3 + x2 + x + 1) = x²(x + 1) + 1(x + 1)

⇒ (x + 1) (x² + 1)

⇒ x⁴ – 1 = (x² – 1) (x² + 1)

⇒ (x + 1) (x – 1) (x² + 1)

∴ Required HCF is (x + 1) (x2 + 1)

GIVEN:

The product of two numbers is 1521 and the HCF of these numbers is 13.

CONCEPT:
HCF: The highest common factor (HCF) is found by finding all common factors of two numbers and selecting the largest one.

CALCULATION:

Suppose the numbers are 13a and 13b as the HCF of these numbers is 13.

We can write:

13a × 13b = 1521

⇒ ab = 9

∴ Only possible pair is 13, 117

Mistake PointsAccording to question,

ab = 9

For a = 1 and b = 9 

The numbers will be 13 and 117 and their HCF will be 13

Here we will not consider a = 3 and b = 3.

The numbers will be 39 and 39.

Here HCF would be 39 which does not satisfy the given condition.

Given: 

The sum of two number positive number is 240 and their HCF is 15.

Calculation: 

Let two positive number is 15x and 15y where x and y should be coprime that means x and y should have HCH as 1.

According to the question

The sum of the number is 

⇒ 15x + 15y = 240

⇒ x + y = 16

Now, we have to find the number of pair in which sum of the two number is 16 but no common factor between them, such pair is

⇒ (1, 15) (3, 13) (5, 11) (7, 9)

∴ Total possible pairs is 4.

Confusion Points

We can't take (2, 14), (4, 12), (6, 10), (8, 8) Because In these cases the pair should be co-prime.

Given:

The product of the two numbers is 2160 and their HCF is 12.

Calculation:

Let the number be x and y.

According to the question,

12x × 12y = 2160

⇒ 144xy = 2160

⇒ xy = \(\dfrac{2160}{144}\) = 15

⇒ xy = 15

Possible pair = (1 × 15) and (3 × 5)

∴ The number of possible pairs of such numbers is 2.

Given:

HCF of two number is 4 and the sum of those two numbers is 36.

Concept Used:

Concept of HCF

HCF is the least common factor among two or more numbers.

Calculation:

HCF of two number is 4

Let, those numbers be 4x and 4y where x and y are prime to each other

Accordingly,

4x + 4y = 36

⇒ 4(x + y) = 36

⇒ (x + y) = 9

Now,

9 = 8 + 1

9 = 7 + 2

9 = 6 + 3

9 = 5 + 4

In all these cases only (8,1); (7,2); and (5,4) are prime to each other. So, such three pair is possible

∴ Such three pair of number is possible.

Given

Ratios of numbers: 4 : 5 : 6

LCM of the numbers: 180

Concept:

When three numbers are in the ratio a:b:c, and their LCM is n, then their HCF is n/(abc).

Solution:

HCF of the numbers = 180/LCM of (4, 5 and 6) = 180/60 ⇒ 3

Therefore, their HCF is 3.
  Alternate Method

Ratio of numbers = 4 : 5 : 6

Let numbers be 4x, 5x and 6x where x is the HCF of numbers.

So

LCM of (4x, 5x, 6x) = 60x

60x = 180

so

x = 3 = HCF of numbers

So 3 is the correct answer.